waecmaths question:
Q is 32km away from P on a bearing of 042o and R is 25km from P on a bearing of 132o. Calculate the bearing of R from Q
Option A:
122o
Option B:
184o
Option C:
190o
Option D:
226o
waecmaths solution:
$\begin{align} & \tan \theta =\frac{\left| PR \right|}{\left| PQ \right|}=\frac{25}{32} \\ & \theta ={{\tan }^{-1}}(\tfrac{25}{12})={{38}^{\circ }} \\ & \angle PQA=\angle BPQ={{42}^{\circ }} \\ & \angle RQA=\angle PQA-\angle PQR \\ & \angle RQA={{42}^{\circ }}-{{38}^{\circ }}={{4}^{\circ }} \\\end{align}$The bearing of R of Q is 180o + 4o =184o
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